Question: For what real value of $v$ is $\frac{-21-\sqrt{301}}{10}$ a root of $5x^2+21x+v$?
Explanation: We could substitute $(-21-\sqrt{301})/10$ for $x$ in the equation, but the quadratic formula suggests a quicker approach. Substituting $5$, $21$, and $v$ into the quadratic formula gives  \[
\frac{-(21)\pm\sqrt{(21)^2-4(5)(v)}}{2(5)}= \frac{-21\pm\sqrt{441-20v}}{10}.
\]Setting $(-21+\sqrt{441-20v})/10$ and $(-21-\sqrt{441-20v})/10$ equal to $(-21-\sqrt{301})/10$, we find no solution in the first case and $441-20v=301$ in the second case.  Solving yields $v=(301-441)/(-20)=(-140)/(-20)=\boxed{7}$.